Laplace transform, pretty quick (renewal process)

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  1. wanderlust.xx's Avatar
    1 23-04-2012  11:46
    • TSR Demigod
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    Laplace transform, pretty quick (renewal process)
    I have a renewal function given as m(t) = \dfrac{1}{4}(1-e^{-2\lambda t} +6\lambda t) for \lambda>0.

    The interarrival times are identically distributed with distribution density function f(t) which is given by the equation f''(t) + 4\lambda f'(t) +3 \lambda^2 f(t) = 0, \ t>0.

    I need to show that f(t) = \dfrac{1}{2}\lambda (e^{-\lambda t} + 3 e^{\lambda t})

    Having solved the second order differential equation, I get that f(t) = A e^{- \lambda t} + Be^{-3 \lambda t}. Clearly I just need my constants A and B, but the only possible way I can do it straight from here is by using the fact that f is a density, so \int f dt = 1. I'm not sure what to do from here.

    I know that \bar{m}(s) = \dfrac{\bar{f}(s)}{s(1-\bar{f}(s))}, where \bar{g}(s) = \int_0^{\infty} e^{-st}g(t) dt.

    So I found \bar{m}(s) = \dfrac{1}{4} \left(\dfrac{1}{s} - \dfrac{1}{s+2\lambda} + \dfrac{6\lambda}{s^2} \right)

    Now rearranging the aforementioned equation, I'll get \bar{f} = \dfrac{s\bar{m}}{1+s\bar{m}}... but substitution is really tricky. I expect that I've gone wrong somewhere in my logic but I can't see why.

    Edit: I guess I could just use the f(t) I'm trying to prove, find its transform and show that it gives us \bar{m}(s) = \dfrac{1}{4} \left(\dfrac{1}{s} - \dfrac{1}{s+2\lambda} + \dfrac{6\lambda}{s^2} \right)...? But this is really dodgy!

    Anyone have any suggestions?
    Last edited by wanderlust.xx; 23-04-2012 at 13:33.
  2. Glutamic Acid's Avatar
    2 23-04-2012  13:27
    • TSR Legend
    • Location: E.I.R.E. / S.E. / Cam
    Re: Laplace transform, pretty quick (renewal process)
    (Original post by wanderlust.xx)
    I have a renewal function given as m(t) = \dfrac{1}{4}(1-e^{-2\lambda t} +6\lambda t) for \lambda>0.

    The interarrival times are identically distributed with distribution function f(t) which is given by the equation f''(t) + 4\lambda f'(t) +3 \lambda^2 f(t) = 0, \ t>0.

    I need to show that f(t) = \dfrac{1}{2}\lambda (e^{-\lambda t} + 3 e^{\lambda t})

    Having solved the second order differential equation, I get that f(t) = A e^{- \lambda t} + Be^{-3 \lambda t}. Clearly I just need my constants A and B, but the only possible way I can do it straight from here is by using the fact that f is a density, so \int f dt = 1. I'm not sure what to do from here.
    f isn't a density, it's a distribution function (at least that's what's written). Unfortunately this doesn't make sense. Have you quoted the question verbatim?
  3. wanderlust.xx's Avatar
    3 23-04-2012  13:31
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    • Location: London
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    Re: Laplace transform, pretty quick (renewal process)
    (Original post by Glutamic Acid)
    f isn't a density, it's a distribution function (at least that's what's written). Unfortunately this doesn't make sense. Have you quoted the question verbatim?
    Ah yes, sorry:

    "Suppose that the common distribution function F of the interarrival times is repeatedly differentiable for t>0 and that f(t) = F'(t) is the common density function for t>0 of the interarrival times."

    So it is indeed a density.
  4. Glutamic Acid's Avatar
    4 23-04-2012  13:44
    • TSR Legend
    • Location: E.I.R.E. / S.E. / Cam
    Re: Laplace transform, pretty quick (renewal process)
    (Original post by wanderlust.xx)
    Ah yes, sorry:

    "Suppose that the common distribution function F of the interarrival times is repeatedly differentiable for t>0 and that f(t) = F'(t) is the common density function for t>0 of the interarrival times."

    So it is indeed a density.
    I don't think there's anything wrong with taking the Laplace transform of f and solving for A and B. You can substitute in a value of s which may make it easier (this, together with f being a density should give you what you need).
    Last edited by Glutamic Acid; 23-04-2012 at 13:46.
  5. wanderlust.xx's Avatar
    5 23-04-2012  14:05
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    • Location: London
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    Re: Laplace transform, pretty quick (renewal process)
    (Original post by Glutamic Acid)
    I don't think there's anything wrong with taking the Laplace transform of f and solving for A and B. You can substitute in a value of s which may make it easier (this, together with f being a density should give you what you need).
    Hmm, I'm still hopelessly stuck.

    Using the fact that f is a density, I've arrived at \dfrac{A}{\lambda} + \dfrac{B}{3\lambda} = 1.

    So 3A+B = 3\lambda \ ... \ (i)

    The laplace transform of the f I'm trying to find is

    \bar{f}(s) = \dfrac{2\lambda}{s+\lambda}

    Substituting a value of 0 gives that \bar{f}(0) = 2.

    Using the value of f I obtained from the differential equation, my transform was \bar{f}(s) = \dfrac{A}{s+\lambda} + \dfrac{B}{s+ 3\lambda}

    But I don't see how we can use this to solve (i).

    Oh wait, subbing 0 in should yield 1 since (i) is definitely correct, shouldn't it? Eurrghhh.
    Last edited by wanderlust.xx; 23-04-2012 at 14:09.
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